Задача Д21 вариант №8

21 ( 8)

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1-5 Pi (i =

1, 5).

q1

q2

.

, .

3, 4, 5 .

: m1 = m, m2 = 2m, m3 = 2m, m4

= 2m,

m5 = 2m,

f ,

q1 = x,

q2

= ξ.

0 = 0, ξ0

= ξ0 .

: q10 = 0, q20 = 0,

q10 = 0, q20

= ξ0

0 = 0,

ξ0 = 0,

d

∂T -

∂T = -

∂ + Q ;

(1)

dt ¶x

¶x

¶

1

d

∂T -

∂T = -

∂ + Q .

(2)

dt ¶ξ

¶ξ

¶ξ

2

# – $" ; – %$";

Q1 Q2 – , & .

5

T = i .

(3)

x

x

ξ

= x,

ω 5 =

ω4 =

v4 = 0.5 ,

v3 = 0.5

, v1 = 0.5 + ξ ,

v2

,

,

ω 3 = r .

(4)

r

2r

% % :

J3 = J4 = J5

=

2m

r 2

= mr 2 .

(5)

2

2

2

3

2

2

2

+ 0.5× mr

2

ξ

2

);

= 0.5m3v3

+ 0.5J3ω3

= 0.5× 2m ×(0.5x)

×

= 0.5m ×(0.5x

+ ξ

r

2

2

2

2

2

x

2

4

= 0.5m4v4

+ 0.5J4ω4

+ 0.5× mr

×

;

= 0.5× 2m ×(0.5x)

2r

= 0.375mx

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2

2

2

x

2

5 = 0.5J5ω5

= 0.5

× mr

×

.

= 0.5mx

r

( (3),

2

2

2

2

2

2

2

2

). (6)

T = 0.5m(0.5x + ξ )

+ mx

+ 0.5m(0.5x

+ ξ

) + 0.375mx

+ 0.5mx

= m(2.25x

+ 0.5xξ + ξ

= 1 + 2 + 3 + 4 .

(7)

= -m g(x + ξ ) = -mg(x + ξ );

= 2mgxsin 45 =

2

2mgx;

1

1

3 = -m3 g ×0.5x = -2mg ×0.5x = -mgx;

4

= -m4 g ×0.5x = -2mg ×0.5x = -mgx.

( (7),

= -mg(x + ξ ) +

mgx - mgx - mgx = -mg((3 -

)x + ξ ).

(8)

2

2

Q =

N1

;

Q

=

N2

.

1

x

2

ξ

Q = -m

gf cos 45

= -2mgf 0.5

= -

Q

= 0.

(9)

'

2

2mgf ;

2

1

2

((6) (7) (9) (1) (2),

∂T

d ∂T

∂T

∂

+ 0.5ξ );

+ ξ );

= 0;

= -mg(3 - 2)

= m(4.5x

= 0.5m(9x

¶x

¶

¶x

dt ¶x

∂T

d ∂T

∂T

∂

+ 2ξ );

+

4ξ );

¶ξ

= 0;

¶ξ

= -mg;

¶ξ

dt ¶ξ

= m(0.5x

= 0.5m(x

0.5m(9x + ξ ) = mg(3 -

) -

9x + ξ = (6 - 2

(1+ f ))g;

(10)

2

2mgf

2

(11)

0.5m(x + 4ξ ) = mg

x + 4ξ = 2g.

(10) (11),

9x + ξ = (6 - 2

(1+ f ))g;

35ξ = (12 + 2

(1+ f ))g;

2

2

9x + 36ξ = 18g.

0.25x + ξ = 0.5g.

12 + 2

(1+ f )

22 - 8

(1+ f )

=

2

g;

=

2

g;

ξ

x

35

35

= 0.5A;

,

= 0.5B.

12 + 2

2(1+ f )

+ 2

2(1+ f )

g = 0.5g

=

12

g.

ξ

0.25x +

35

ξ

35

" = 44 -16

(1+ f ) g;

B = 24 + 4

(1+ f ) g.

2

2

x = 0.5At + C1;

35

x0 = C1

35

x = 0.5At

C1 = 0

x = At 2 + C2

x0 = C2

C2 = 0

x = At 2 .

ξ = 0.5Bt + C3 ;

ξ0

= C3

C3 = ξ0

ξ = 0.5Bt + ξ0

ξ = Bt 2 + ξ0t + C4

ξ0 = C4

C4 = 0 ξ = Bt 2 + ξ0t.

44 -16

(1+ f ) gt 2 ;

ξ = 24 + 4

(1+ f ) gt 2 + ξ0t.

:

x =

2

2

35

35